Recent content by QuestForInsight

By the way, this limit is the Riemann sum of x² over [0, 1]: $\displaystyle \lim_{n \to \infty}\frac{1}{n^3}\sum_{i=1}^{n-1}i^2 = \lim_{n \to \infty}\frac{1}{n^3}\sum_{i=1}^{n}i^2 = \int_{0}^{1}x^2\;{dx} = \frac{1}{3}.$

They have replaced i with i+1. It's sort of like a substitution for sums. Suppose we let k = i-1, then k = 0 when i = 1 and n-1 when i = n, thus we have: $\displaystyle \lim_{n \to \infty}\frac{1}{n^3}\sum_{i=1}^{n}(i-1)^2 = \lim_{n \to \infty}\frac{1}{n^3}\sum_{k=0}^{n-1}k^2 $ But k is a...

Using Bernoulli's Inequality we have: $ \begin{aligned} \displaystyle & \hspace{0.5in}\bigg(1+\frac{x}{n}\bigg)^{\frac{k}{n}} \le \bigg(1+\frac{kx}{n^2}\bigg) \le \bigg(1+\frac{x}{n^2}\bigg)^k \\& \iff \prod_{ 0\le k \le n}\bigg(1+\frac{x}{n}\bigg)^{\frac{k}{n}} \le \prod_{ 0\le k \le...

Not much of a mistake, tbf. In this case interchanging the limit and the sum works; it's the justification that's missing. In your rewrite, I'm not sure how you reduced the summand to $\frac{kx}{n^2}$ without interchanging the sum and the limit, though?

Brilliant! (Clapping) It's much easier than the solution I know.

Show that $$ \lim_{n \to \infty} \prod_{1 \le k \le n}\left(1+\frac{kx}{n^2}\right) = e^{x/2}$$

An alternative would be generating functions (which I'm quite fond of). (Inlove)

Become an expert in arithmetic geometry. I'll honestly die happy if that happens. :cool:

I agree with Jameson. The mistakes you're making are common and most people make them when they're learning calculus for the first time. I would say continuous rectification of these mistakes and the algebra that integration and differentiation usually require build one's algebraic skills better...

$ \lfloor{x}\rfloor = \max\{m\in\mathbb{Z}\mid m < x\}$ when $x$ is not an integer, thus: $ \begin{aligned} \lfloor{x}\rfloor+\lfloor{-x}\rfloor & = \max\{m\in\mathbb{Z}\mid m < x\}+\max\{m\in\mathbb{Z}\mid m < -x\} \\& = \max\{m\in\mathbb{Z}\mid 2m < 0\} = \max\{m\in\mathbb{Z}\mid m < 0\} \\&...

Re: Determine the general term for a sequnce Okay sorry. Then it's me who misread the original thread. I thought all the given terms of the concerned sequence were just 8, 12, 18, and 27 (and to be fair the question does appear that way in the original post). My apologies. I should have paid...

Re: Determine the general term for a sequnce I think you misread the thread.

Re: Determine the general term for a sequnce I've noticed that $2(8)-\frac{1}{2}(8) = 12$, $2(12)-\frac{1}{2}(12) = 18$, $2(18)-\frac{1}{2}(18) = 27.$ We wish to find the sequence that satisfies $a_{k+1} = \frac{3}{2}a_{k} ~~ (k \ge 0; ~ a_{0} = 8)$. Define the generating function...

Re: Determine the general term for a sequnce This also works: $\displaystyle f(k) = \frac{3^k}{2^{k-3}} ~~ (k \ge 0)$.

I think you are bit confused here. You're asked to solve x(x-3) = 0. Think about it this way: when do you get 0 from the product of two numbers? When at least one of them is zero! Here you have a product of x and x-3 -- that's x(x-3). You're told that it's zero. It can only be so if x = 0 or x-3...